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Let $f(x)$ be a polynomial of degree $n$ with coefficients in $\mathbb{Q}$. There are well-known ways to construct a $n \times n$ matrix $A$ with entries in $\mathbb{Q}$ whose characteristic polynomial is $f$. My question is: when is it possible to choose $A$ symmetric?

An obvious necessary condition is that the roots of $f$ are all real, but it is not clear to me even in the case $n = 2$ that this is sufficient. In degree $2$ this comes down to determining whether or not every pair $(p,q)$ which satisfies $p^2 > 4q$ (the condition that $x^2 + px + q$ has real roots) can be expressed in the form $$p = -(a + c)$$ $$q = ac - b^2$$ where $a$, $b$, and $c$ are rational. I have some partial results from just fumbling around and checking cases, but it seems clear that a more conceptual argument would be required to handle larger degrees.

Rodrigo de Azevedo
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Paul Siegel
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    in degree $2$, this is possible if and only if $p^2-4q = x^2+y^2$ for some rational numbers $x,y$. So requiring $p^2-4q \ge 0$ is not enough. Also, if you work over $\Bbb Q(i)$ then every degree $2$ polynomial is a characteristic polynomial of a symmetric matrix – mercio Nov 20 '13 at 12:16
  • Cool, I agree. Can you formulate a corresponding number theory problem in higher degrees? – Paul Siegel Nov 21 '13 at 01:50
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    Miroslav Fiedler and Gerhard Schmeisser present a tridiagonal companion matrix (that can be symmetrized if all the roots of the starting polynomial are real); the question then becomes: when will the subdiagonal elements be rational? (Alternatively: is there a similarity transformation that will result in a matrix with rational entries?) – J. M. ain't a mathematician Jun 03 '16 at 17:41
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    There is some literature on these questions. For example, you might look at this article. – Robert Israel Jun 03 '16 at 18:22
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    The paper Edward A. Bender: Characteristic polynomials of symmetric matrices1968 http://dx.doi.org/10.2140/pjm.1968.25.433 has the following abstract: Let $F$ be a field and $p$ an $F$-polynomial. We say that $p$ is $F$-real if and only if every real closure of $F$ contains the splitting field of $p$ over $F$. Our main purpose is to prove Theorem 1. Let $F$ be an algebraic number field and $p$ a monic $F$-polynomial with an odd degree factor over $F$. Then $p$ is $F$-real if and only if it is the characteristic polynomial of a symmetric $F$-matrix. – Martin Sleziak Jun 06 '16 at 08:34

2 Answers2

4

I found an answer to your question in the following paper: Estes, Dennis R.(1-SCA); Guralnick, Robert M.(1-SCA) Minimal polynomials of integral symmetric matrices. Computational linear algebra in algebraic and related problems (Essen, 1992). Linear Algebra Appl. 192 (1993), 83–99.

In the introduction, the authors mention:

"Bender, in a series of papers, studied this problem for $\Bbb Q$, the rationals. He proved that any totally real monic polynomial over $\Bbb Q$ of odd degree can occur as the characteristic polynomial of a symmetric rational matrix. This fails already for polynomials of degree 2. The problem of determining precisely which polynomials can occur as characteristic polynomials of rational symmetric matrices is still not solved. The conjecture is that this is a local problem (i.e., if $f \in \Bbb Q[X]$ is monic of degree $n$, then $f$ is the characteristic polynomial of some element of $S_n(\Bbb Q)$ if and only if $f$ is the characteristic polynomial of some element of $S_n(\Bbb R)$ and $S_n(\Bbb Q_p)$ for all primes $p$."

paf
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Oblomov
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I guess that the following paper shows how, for any given polynomial, it can be determined (calculated) a certain symmetric matrix A for which said polynomial is the characteristic polynomial of A (see link below).

Given the demonstration presented in the paper, I would not say that given a certain polynomial, matrix A could "be chosen", since that would imply that the degrees of freedom allow for such choosing to be possible.

As the demonstration shows, for any given polynomial a certain matrix A can be calculated, which means that there always is a system of linear equations that can be solved (the determinant of the system of linear equations is != 0), so, it means that that the degrees of freedom is always zero, so, there is no way that matrix A could be "chosen", even though the solution of A could be guessed.

Kind regards, GEN

Fiedler, Miroslav, Expressing a polynomial as the characteristic polynomial of a symmetric matrix, Linear Algebra Appl. 141, 265–270 (1990). Zbl 0721.15005.

The Amplitwist
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GEN
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    This paper addresses the problem of constructing a symmetric matrix with real entries whose characteristic polynomial is a prescribed polynomial with real roots. But this is not hard: if $r_1, \ldots r_n$ are the roots then take $A = diag(r_1, \ldots, r_n)$. The point is that this matrix will not in general have entries in $\mathbb{Q}$ even if the coefficients of the polynomial are in $\mathbb{Q}$, and this is not addressed in the paper. (Indeed, the goal of the paper appears to be to construct a symmetric matrix without having direct access to the roots of the polynomial.) – Paul Siegel Jul 05 '16 at 17:08
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    Also, don't get hung up on the word "choose": it is perfectly possible to choose an element from a one element set! – Paul Siegel Jul 05 '16 at 17:10